Topics in Complexity Theory
11 March, 2015
Lec. 13: Monomial Ordering
Lecturer: Neeraj Kayal
Scribe: Vineet Nair
In this lecture we introduce the concept of monomial ordering and look at problems where it is
useful.
Idea: Say we have two univariate polynomials:
f (x) = α0 + α1 x + ... + αd xd
g(x) = β0 + β1 x + ... + βe xe
f (x)g(x) = α0 β0 + (α0 β1 + α1 β0 )x + ... + αd βe xd+e
Claim 13.0.1. Given two univariate polynomials f and g of degree d and e respectively as above,
the polynomial f g has degree d+e.
Proof. We order the monomials in a univariate polynomial as follows: xi > xj if i > j. xi is the
leading monomial in a polynomial h, if xi is greater than all other monomials in h. It is easy to
see that, f is a degree d polynomial iff the leading monomial in f is xd . If the leading monomial
in h1 and h2 is xi and xj then the leading monomial in h1 h2 is xi .xj = xi+j . Hence the leading
monomial in f g is xd+e . Hence degree of f g is d+e.
We wish to generalize the monomial ordering concept used in the above proof for univariate polynomials to multivariate monomials. Frome here on X represents the set of variables
X = {x1 , x2 , ..., xn } ,α represents the vector (α1 , α2 , ..., αn ) and X α represents the monomial
xα1 1 xα2 2 ...xαnn . We want to establish a variable ordering such that the following holds.
1. For any two monomials X α and X β either X α > X β or X α < X β .
2. For all α and β, X α X β > X α .
3. For any monomial X α , there are a finite number of monomials samller than it, i.e {m : X α >
m} is finite.
Examples of variable ordering
1. Pure lexicographic ordering:
In this the variables are ordered as x1 > x2 > ... > xn and for any two degree vectors α =
(α1 , α2 , ..., αn ) and β = (β1 , β2 , ..., βn ) where α 6= β
for α1 6= β1 and α1 > β1 ⇒ X α > X β
else for α2 6= β1 and α2 > β2 ⇒ X α > X β
.
.
.
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else for αn 6= βn and αn > βn ⇒ X α > X β
2. Graded lexicographic ordering
For any two degree vectors α = (α1 , α2 , ..., αn ) and β = (β1 , β2 , ..., βn ) where α 6= β
if
Pn
i=1 αi
else if
Pn
>
Pn
i=1 αi
i=1 βi
=
Pn
then X α > X β
i=1 βi
then use pure lexicographic ordering to order X α , X β .
Leading Monomial and Trailing Monomial:
Given
f=
X
aα X α
α∈Zn
we say a monomial X α is the leading monomial of f , represented as LM(f ) if for all X β > X α ,
aβ = 0. Similarly we can define the trailing monomial of f , represented as TM(f ) if for all X β < X α ,
aβ = 0.
Claim 13.0.2. Given two multivariate polynomials f and g, LM(f g)=LM(f )LM(g).
Proof. Say for contradiction LM(f g)6=LM(f )LM(g). Let X α and X β be the leading monomial of
LM(f ) and LM(g) respectively. Let LM(f g)=X α1 X β1 where X α1 and X β1 are monomials in f
and g respectively. Hence either X α1 < X α or X β1 < X β . W.l.o.g assume X α1 < X α . This
implies α1 < α (since the ordering is implicitly on the degree vectors). Since X α1 X β1 is the leading
monomial, α1 + β1 > α + β. But X α and X β are the leading monomials of LM(f ) and LM(g)
respectively, hence α + β > α1 + β1 . Thus we get a contradiction.
We will see a couple of examples where we can use monomial ordering to solve the problem.
1. Show that the monomial x2 y 3 cannot be expressed as a power of any polynomial, i.e x2 y 3 6= f e
where f is polynomial and e 6= 1.
Say for contradiction x2 y 3 = f e . This implies LM(x2 y 3 )=LM(f e )=LM(f )e . Thus x2 y 3 = xem y en .
Since gcd(2,3)=1 we have e = 1, m = 2 and y = 3. Hence a contradiction. This method also
extends to show x2 y 3 + αt x2 y 2 + ... + α0 x0 is not power of any polynomial.
2.Suppose we have two multivariate polynomials f (X) and g(X) where X = {x1 , x2 , ..., xn } and
we need to determine whether f (X) and g(X) have a common root.
Observe that if f (X) and g(X) have a common root then deg(gcd(f (X), g(X))) > 0. Hence
we perform euclids algorithm on f (X) and g(X) to determine their gcd. Suppose deg(f (X))) >
deg(g(X)). We divide f (X) by g(X). Say we get
f (X) = g(X)q(X) + r(X)
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where q(X) and r(X) are the qoutient and remainder polynomials respectively. We know gcd(f (X), g(X))
= gcd(g(X), r(X)). In multivariate case we need to determine which of the polynomials g(X) or
r(X) is smaller to continue the recursion. r(X) is smaller than g(X) iff LM(r(X)) < LM(g(X)).
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